Example 19
A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.
Given AB = 15 cm
& AP = 6 cm
Now,
PB = AB – AP
PB = 15 – 6
PB = 9cm
Drawing PQ ⊥ BO and PR ⊥ OA
Hence, PQ = x & PR = y
Let ∠ PAR = θ
Now, PQ & AO are parallel lines
(As both are perpendicular to y-axis)
& BA is the transversal
So ∠ BPQ = ∠ PAR = θ
In right triangle ∆ BPQ
cos𝜃 = 𝐵𝑎𝑠𝑒/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos𝜃 = 𝑃𝑄/𝐵𝑃
𝒄𝒐𝒔𝜽 = 𝒙/𝟗
In right triangle ∆ PAR
〖 sin〗𝜃 = (ℎ𝑒𝑖𝑔ℎ𝑡 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
〖 sin〗𝜃 = 𝑃𝑅/(𝐴𝑃 )
〖 𝒔𝒊𝒏〗𝜽 = 𝒚/𝟔
Now we know that,
sin2𝜃 + cos2𝜃 = 1
Putting 〖 𝑠𝑖𝑛〗𝜃 = 𝑦/6 and 𝑐𝑜𝑠𝜃 = 𝑥/9
(𝑦/6)^2 + (𝑥/9)^2 = 1
𝑦2/36 + 𝑥2/81 = 1
𝒙𝟐/𝟖𝟏 + 𝒚𝟐/𝟑𝟔 = 1
Hence it satisfies the equation of ellipse
𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1
Thus, locus of P is ellipse
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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